Question

one ticket is selected at ramdom from 50 numbered 00,01,02....49. Then the probability that the sum of the digit on the selected tucket is 8,given that the product of these digit is zero

Answer 1

E:Tickets in which sum of the digit is eight
F:product of digits is zero
n(E) :08,17,26,35,44,
P(E) =5/50
=1/10
n(F):08
P(F)=1/50
P(EintersectionF) =1/10
So reqd probability
P(E/F)=p(EintersectionF)/p(F)
=1/10÷1/50
=1/5
=0.2

Answer 2

Answer:

Required probability is $\frac{1}{14}$.

Step-by-step explanation:

$E_1$ = Event that the product of the digits is zero.

$E_2$ = Event that the sum of the digits is 8.

Favourable cases to :-

$E_1$ = {00,01,02,03,...,09,10,20,30,40}

$E_2$ = {08,17,26,35,44}

$n$($E_1$ ∩ $E_2$) = 1

∴ Required probability :-

$P\frac{E_2}{E_1}$ = $n\frac{E_1unionE_2}{n(E_1)}$ = $\frac{1}{14}$